Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}3x-6y &= 2 \\ 6x+8y &= -1\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $8y = -6x-1$ Divide both sides by $8$ to isolate $y$ $y = {-\dfrac{3}{4}x - \dfrac{1}{8}}$ Substitute this expression for $y$ in the first equation. $3x-6({-\dfrac{3}{4}x - \dfrac{1}{8}}) = 2$ $3x + \dfrac{9}{2}x + \dfrac{3}{4} = 2$ Simplify by combining terms, then solve for $x$ $\dfrac{15}{2}x + \dfrac{3}{4} = 2$ $\dfrac{15}{2}x = \dfrac{5}{4}$ $x = \dfrac{1}{6}$ Substitute $\dfrac{1}{6}$ for $x$ back into the top equation. $3( \dfrac{1}{6})-6y = 2$ $\dfrac{1}{2}-6y = 2$ $-6y = \dfrac{3}{2}$ $y = -\dfrac{1}{4}$ The solution is $\enspace x = \dfrac{1}{6}, \enspace y = -\dfrac{1}{4}$.